Why is mysqli_query failing - returning object when it should return resource

Question

I have tried everything to find out why mysqli_query is failing. Can anyone shed light as to what I'm doing wrong. would be possible that I am not connecting to the dababase anymore??!! Thank you in advance!

  function email_exists($email){
      $email = sanitize($email);
    $db = new mysqli('localhost','root','','secured_login');
    if($db->connect_errno){
        $connect_error = 'Sorry, we are experiencing connection problems.'; 
        die ($connect_error);
    }
    return (mysql_result(mysqli_query($db, "SELECT COUNT(`user_id`) FROM `users` WHERE `email` = '$email'"), 0) == 1) ? true : false;
}

error

Warning: mysql_result() expects parameter 1 to be resource, object given in....

Alternative solution using mysqli_fetch_row(); <----- Isthe below alternative valid?

  function email_exists($email){
    $email = sanitize($email);
    $db = new mysqli('localhost','root','','secured_login');
        if($db->connect_errno){
            $connect_error = 'Sorry, we are experiencing connection problems.'; 
            die ($connect_error);
        }
    $query = "SELECT COUNT(`user_id`) FROM `users` WHERE `email` = '$email'";
    if ($result = mysqli_query($db, $query)){
        while ($result= mysqli_fetch_row($result)){
            return ($result);
        }           
    }
}

Any feedback is appreciated!

score 2 · Accepted Answer · edited May 23 '17 at 12:06

2

At it's simplest form, you should be looking at something similar to the following,

function email_exists($email){
    $email = sanitize($email);
    $db = new mysqli('localhost','root','','secured_login');
    if($db->connect_errno){
        $connect_error = 'Sorry, we are experiencing connection problems.'; 
        die ($connect_error);
    }
    $query = $db->query("SELECT `user_id` FROM `users` WHERE `email` = '$email'");
    return ($query->num_rows > 1) ? true : false;
}

Remember to sanitize your inputs, or even better, use prepared statements.

edited May 23 '17 at 12:06

Community

1
1

answered Jan 27 '14 at 12:31

Ben Fortune

31,623
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good, but you should explain why instead of blindly offering a solution. Also, explaining about `mysqli_fetch_*` may be a good idea. – Carlos Campderrós Jan 27 '14 at 12:32
There is allot to learn about php. Thank you for your input! – user3001162 Jan 27 '14 at 12:38

score 0 · Answer 2 · answered Jan 27 '14 at 11:54

0

You're mixing up mysql with mysqli, why?

I presume it's a typo - but change mysql_result to mysqli_result

answered Jan 27 '14 at 11:54

Ryan

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because I tried changing to that but I get the following error message ------Fatal error: Call to undefined function mysqli_result() – user3001162 Jan 27 '14 at 11:55
Should I use mysqli_fetch_assoc? any feedback? – user3001162 Jan 27 '14 at 12:12
1

That's because [mysqli_result](http://www.php.net/manual/en/class.mysqli-result.php) is a class, not a function. – Ben Fortune Jan 27 '14 at 12:21

Why is mysqli_query failing - returning object when it should return resource

2 Answers2